If the sum of first $n$, $2n$ and $3n$ terms of an A.P be $S_1$, $S_2$ and $S_3$ respectively, then prove that $\;$ $S_3 = 3 \left(S_2 - S_1\right)$
Let the first term of A.P $= a$
Let the common difference of A.P $= d$
Sum of first $n$ terms of A.P $= S_1 = \dfrac{n}{2} \left[2a + \left(n - 1\right)d\right]$ $\;\;\; \cdots \; (1)$
Sum of first $2n$ terms of the A.P $= S_2 = \dfrac{2n}{2} \left[2a + \left(2n - 1\right)d\right]$
i.e. $\;$ $S_2 = n \left[2a + \left(2n - 1\right)d\right]$ $\;\;\; \cdots \; (2)$
Sum of first $3n$ terms of the A.P $= S_3 = \dfrac{3n}{2} \left[2a + \left(3n - 1\right)d\right]$ $\;\;\; \cdots \; (3)$
From equations $(1)$ and $(2)$,
$\begin{aligned}
S_2 - S_1 & = n \left[2a + \left(2n - 1\right)d\right] - \dfrac{n}{2} \left[2a + \left(n - 1\right)d\right] \\\\
& = n \left\{2a + 2nd - d - \dfrac{1}{2} \left(2a + nd - d\right) \right\} \\\\
& = n \left\{2a + 2nd - d - a - \dfrac{nd}{2} + \dfrac{d}{2} \right\} \\\\
& = n \left\{a + \dfrac{3nd}{2} - \dfrac{d}{2} \right\} \\\\
& = \dfrac{n}{2} \left[2a + \left(3n - 1\right)d\right]
\end{aligned}$
$\therefore \;$ $3 \left(S_2 - S_1\right) = \dfrac{3n}{2} \left[2a + \left(3n - 1\right)d\right] = S_3$ $\;\;\;$ [by equation $(3)$]
i.e. $\;$ $S_3 = 3 \left(S_2 - S_1\right)$
Hence proved.